Question

1.已知（a-c)^2-4(a-b)(b-c)=0,求证2b=a+c

Answer 1

1.已知（a-c)?-4(a-b)(b-c)=0,求证2b=a+c因为（a-c)?-4(a-b)(b-c)=0故：-2ac+c?-4ab+4ac+4b?-4bc=0即：+c?+4b?+2ac-4ab-4bc=0即：（a+c-2b）?=0故：a+c-2b=0故：2b=a+c 2.如果a?+b?=1,+d?=1,且ac+bd=0,试证明ab+cd=0因为a?+b?=1,+d?=1故：=1-b?,=1-c?因为ac+bd=0故：ac=-bd故：=b?故：(1-b?)c?=b?(1-c?)故：=b?同理：=d?故：=c?因为ac+bd=0,对于ac、bd中一正一负,即：a、b、c、d中有奇数个负数（一正三负、三正一负）或全等于0故：ab=-cd故：ab+cd=0 3.计算(a-1) (a?+a+1) (a+1) (a?-a+1) =[(a-1) (a?+a+1) [ (a+1) (a?-a+1)] =（a?-1）?(a?+1) =[（a?-1）(a?+1) =（a^6-1）?=a^12-2a^6+1 4.已知a+b+c=4,ab+bc+ac=4,求a?+b?+c?因为a+b+c=4故：(a+b+c) =4?故：+b?+c?+2ab+2bc+2ac=16因为ab+bc+ac=4故：+b?+c?=8